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Lesson 3.2 Exercises, pages 190–195
Students should verify all the solutions.
A
4. Which equations are quadratic equations? Explain how you know.
a) 3x2 = 30
b) x2 - 9x + 8 = 0
This equation is quadratic
because the variable term
with the greatest exponent
is an x 2-term.
c) x3 - x2 + 5 = 0
This equation is not
quadratic; it cannot be written
in the form ax 2 bx c 0
because it has one x 3-term.
This equation is quadratic
because it is written in the
form ax 2 bx c 0.
d) 6x + 5 = x - 7
This equation is not quadratic;
it cannot be written in the form
ax 2 bx c 0 because it
does not have an x 2-term.
5. Solve each quadratic equation. Verify the solutions.
a) (x + 5)(x + 8) = 0
Either x 5 0, then
x 5; or x 8 0,
then x 8
c) (2x - 3)(x + 6) = 0
Either 2x 3 0
2x 3
x 1.5
or x 6 0, then x 6
b) (x - 1)(x - 10) = 0
Either x 1 0, then x 1;
or x 10 0, then x 10
d) (3x + 2)(x - 5) = 0
Either 3x 2 0
3x 2
x
2
3
or x 5 0, then x 5
6. Solve.
a) 4(x + 5)(x + 9) = 0
Either x 5 0, then
x 5; or x 9 0,
then x 9
c) x(x - 4) = 0
8
b) 3x(x + 4) = 0
Either 3x 0, then x 0;
or x 4 0, then x 4
d) 5(2x - 1)(3x + 7) = 0
1
2
Either x 0;
Either 2x 1, then x ;
or x 4 0, then x 4
or 3x 7 0, then x 3.2 Solving Quadratic Equations by Factoring—Solutions
7
3
DO NOT COPY.
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B
7. Solve by factoring. Verify the solutions.
a) x2 - 6x + 5 = 0
b) 3x2 - 21x - 54 = 0
(x 1)(x 5) 0
Either x 1 0, then x 1;
or x 5 0, then x 5
c) 2x2 - 15x + 25 = 0
3(x2 7x 18) 0
3(x 9)(x 2) 0
Either x 9 0, then x 9;
or x 2 0, then x 2
d) 10x2 + x - 3 = 0
(2x 5)(x 5) 0
(2x 1)(5x 3) 0
Either 2x 5 0,
then x 2.5;
or x 5 0, then x 5
Either 2x 1 0, then x ;
or 5x 3 0, then x 1
2
3
5
8. Solve by factoring.
a) x2 - 6x = 27
b) 3x2 - 4x = 7
x2 6x 27 0
(x 9)(x 3) 0
3x 2 4x 7 0
(3x 7)(x 1) 0
Either x 9 0, then x 9;
Either 3x 7 0, then x ;
or x 3 0, then x 3
or x 1 0, then x 1
c) x2 - 8x + 12 = 12
7
3
d) 3x2 - 6x = 105
x2 8x 0
x(x 8) 0
Either x 0;
or x 8 0, then x 8
3x2 6x 105 0
3(x2 2x 35) 0
3(x 7)(x 5) 0
Either x 7 0, then x 7;
or x 5 0, then x 5
9. A student wrote the solution below to solve this quadratic equation:
(x - 5)(x + 2) = 8
Either x - 5 = 2
x = 7
or
x + 2 = 4
x = 2
Identify the error, then write the correct solution.
One side of the equation must be 0 before factoring.
(x 5)(x 2) 8
x2 3x 10 8 0
x2 3x 18 0
(x 3)(x 6) 0
Either x 3 0
or
x60
x 3
x6
©P
DO NOT COPY.
3.2 Solving Quadratic Equations by Factoring—Solutions
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10. Solve each equation.
a) (x + 3)(x + 4) = 6
b) x2 - 9 = 4x + 36
x2 7x 12 6 0
x2 7x 6 0
(x 1)(x 6) 0
Either x 1 0,
then x 1;
or x 6 0, then x 6
c) 3x2 + 6 = x(x + 13)
x2 9 4x 36 0
x2 4x 45 0
(x 5)(x 9) 0
Either x 5 0,
then x 5;
or x 9 0, then x 9
d) 2x(x - 6) + 3x = 2x - 9
3x2 6 x2 13x
2x 13x 6 0
(2x 1)(x 6) 0
Either 2x 1 0,
then x 0.5;
or x 6 0, then x 6
2x2 12x 3x 2x 9 0
2x2 11x 9 0
(2x 9)(x 1) 0
Either 2x 9 0,
then x 4.5;
or x 1 0, then x 1
2
11. Create a quadratic equation that has exactly one root. Explain your
strategy.
Sample response: Work backward. Choose a root of x 6.
The equation is: (x 6)(x 6) 0
Expand:
x2 12x 36 0
12. Solve each equation, then verify the solution:
a)
√
23 - x = x - 3
√
( 23 x)2 (x 3)2
b)
23 x x2 6x 9
0 x2 5x 14
0 (x 7)(x 2)
Either x 7 0, then x 7;
or x 2 0, then x 2
I used mental math to verify.
x 2; the root is x 7
√
2x2 + 1 + 1 = 2x
√ 2
2x 1 2x 1
√ 2
( 2x 1)2 (2x 1)2
2x2 1 4x2 4x 1
0 2x2 4x
0 2x(x 2)
Either 2x 0, then x 0;
or x 2 0, then x 2
I used mental math to verify.
x 0; the root is x 2
x2
7x
13. Solve this equation: 2 + 6 = 1
Multiply each term by the common denominator 6.
3x2 7x 6
2
3x 7x 6 0
(3x 2)(x 3) 0
Either 3x 2 0
or
x30
x
10
2
3
x 3
3.2 Solving Quadratic Equations by Factoring—Solutions
DO NOT COPY.
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14. A football is kicked vertically. The approximate height of the
football, h metres, after t seconds is modelled by this formula:
h = 1 + 20t - 5t2
a) Determine the height of the football after 2 s.
h 1 20t 5t2
Substitute t 2.
h 1 20(2) 5(2)2
h 21
The football is 21 m high.
b) When is the football 16 m high?
h 1 20t 5t2 Substitute h 16, then solve for t.
16 1 20t 5t2
2
5t 20t 15 0
5(t2 4t 3) 0
5(t 1)(t 3) 0
So, t 1 or t 3
The football is 16 m high after 1 s and after 3 s.
c) Why are there two solutions for part b?
The football is 16 m high as it is rising and as it is falling.
15. A student wrote the solution below to solve this quadratic equation:
5x2 = 25x
25x
25x2
=
5x
5x
x = 5
Identify the error, then write the correct solution.
The student should not have divided by x since x 0 is a solution.
5x2 25x
2
5x 25x 0
5x(x 5) 0
Either 5x 0
or
x50
x0
x5
The roots are: x 0 and x 5
16. When twice a number is subtracted from the square of the number,
the result is 99. Determine the number.
Let x represent the number.
An equation is: x2 2x 99
x2 2x 99 0
(x 11)(x 9) 0
Either x 11 0
or
x 11
The numbers are 11 and 9.
©P
DO NOT COPY.
x90
x 9
3.2 Solving Quadratic Equations by Factoring—Solutions
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17. Cody has a daily reading program where he reads for 2 min more
than he did the day before. On the first day, he read for 10 min.
Cody continued his program until he had read for a total of 6 h.
a) Write an equation to represent the number of days, n, for which
Cody read.
The time, in minutes, for which Cody reads is:
10 12 14 . . .
This is an arithmetic series with 1st term 10 and common difference 2.
The total time Cody reads, 6 h or 360 min, is the sum of this series.
n[2t1 d(n 1)]
2
n[2(10) 2(n 1)]
360 2
Use: Sn 720
720
720
0
0
n(20 2n 2)
n(18 2n)
18n 2n2
2n2 18n 720
n2 9n 360
Substitute: Sn 360, t1 10, d 2
Divide each term by 2.
b) Solve the equation for n.
n2 9n 360 0
(n 24)(n 15) 0
Either n 24 0
or
n 15 0
n 24
n 15
Since the number of days cannot be negative, Cody read for 15 days.
C
18. Solve each equation, then verify the solutions.
a) (2x + 1)2 = (x + 5)2
4x2 4x 1 3x2 6x 24 x2 2x 8 (x 4)(x 2) Either x 4 x
x2 10x 25
Divide each term by 3.
0
0
0
0
or
x20
4
x 2
b) (2x - 1)2 - 2(2x - 1) - 8 = 0
This equation has the same form as the simplified equation in part a,
where x is replaced with 2x 1.
[(2x 1) 4][(2x 1) 2] 0
(2x 5)(2x 1) 0
Either 2x 5 0
or
2x 1 0
x 2.5
x 0.5
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3.2 Solving Quadratic Equations by Factoring—Solutions
DO NOT COPY.
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19. The area of a rectangular sheep pen is 96 m2. The pen is divided
into two smaller pens by inserting a fence parallel to the width of
the pen. A total of 48 m of fencing is used. Determine the
dimensions of the pen.
w
The area of the rectangular pen is 96 m2, and the width is
96
w metres, so the length, in metres, is w .
The total length of fencing is 48 m, so an equation is:
3w 2 aw b 48
96
Multiply each term by w.
3w2 2(96) 48w
3w 48w 192 0 Divide each term by 3.
w2 16w 64 0
(w 8)(w 8) 0
w80
w8
96
The pen is 8 m wide, so its length is
m, or 12 m.
2
8
20. A rectangular garden has dimensions 3 m by 4 m. A path is built
around the garden. The area of the garden and path is 6 times as
great as the area of the garden. What is the width of the path?
Let the width of the path be x metres.
Then the width of the rectangle formed by the garden and path, in
metres, is 3 2x, and the length of this rectangle, in metres, is 4 2x.
The area of the garden is 12 m2.
The areas of the garden and path, in square metres, is:
(3 2x)(4 2x)
An equation is: (3 2x)(4 2x) 6(12)
12 14x 4x2 72
4x2 14x 60 0
Divide each term by 2.
2x2 7x 30 0
(2x 5)(x 6) 0
Either 2x 5 0
or
x60
x 2.5
x 6
Since the width of the path cannot be negative, the width is 2.5 m.
©P
DO NOT COPY.
3.2 Solving Quadratic Equations by Factoring—Solutions
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